Answer:
[tex]\tan{\theta} = \frac{\sqrt{11}}{11}[/tex]
Step-by-step explanation:
Cosecant:
The cosecant is one divided by the sine. Thus:
[tex]\csc{\theta} = \frac{1}{\sin{\theta}}[/tex]
Tangent is sine divided by cosine, so we first find the sine, then the cosine, to find the tangent.
Sine and cosine:
[tex]\sin{\theta} = \frac{1}{\csc{\theta}} = \frac{1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{6}[/tex]
[tex]\sin^{2}{\theta} + \cos^{2}{\theta} = 1[/tex]
[tex]\cos^{2}{\theta} = 1 - \sin^{2}{\theta}[/tex]
[tex]\cos^{2}{\theta} = 1 - (\frac{\sqrt{3}}{6})^2[/tex]
[tex]\cos^{2}{\theta} = 1 - \frac{3}{36}[/tex]
[tex]\cos^{2}{\theta} = \frac{33}{36}[/tex]
First quadrant, so the cosine is positive. Then
[tex]\cos^{2}{\theta} = \sqrt{\frac{33}{36}} = \frac{\sqrt{33}}{6}[/tex]
Tangent:
Sine divided by cosine. So
[tex]\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}} = \frac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{33}}{6}} = \frac{\sqrt{3}}{\sqrt{33}} = \frac{\sqrt{3}}{\sqrt{3}\sqrt{11}} = \frac{1}{\sqrt{11}} \times \frac{\sqrt{11}}{\sqrt{11}} = \frac{\sqrt{11}}{11}[/tex]
The answer is:
[tex]\tan{\theta} = \frac{\sqrt{11}}{11}[/tex]
